How do you integrate int x^3 ln x^4 dx using integration by parts?

2 Answers
Mar 1, 2018

intx^3ln(x^4)dx=x^4ln(x)-1/4x^4+C

Explanation:

First, let's rewrite ln(x^4) as 4ln(x), as ln(x^a)=aln(x)

intx^3ln(x^4)dx=int4x^3ln(x)dx=4intx^3ln(x)dx (We can factor out 4, it's just a constant.)

Make the following choices for u and dv:

u=ln(x)

dv=x^3dx

Thus, du and v become:

du=1/xdx

dv=intx^3dx=1/4x^4 (We're not going to include the constant just yet. It can wait until the end.)

Now it becomes apparent why we made the choices we did for u and v. If we chose dv=ln(x), we would have to integrate ln(x) for v, and would run into our original problem.

Plug in relevant values:

uv-intvdu=1/4x^4ln(x)-int1/4x^4/xdx

We can simplify the integral here:

1/4x^4ln(x)-int1/4x^4/xdx=1/4x^4ln(x)-1/4intx^(cancel(4)3)/cancelx

Thus, we have

1/4x^4ln(x)-1/4intx^3dx=1/4x^4ln(x)-(1/4)(1/4)x^4+C=1/4x^4ln(x)-1/16x^4+C

Let's not forget the 4 we originally factored out...

4(1/4x^4ln(x)-1/16x^4+C)=x^4ln(x)-1/4x^4+C

C remains unchanged; a constant multiplied by a constant remains a constant.

So,

intx^3ln(x^4)dx=x^4ln(x)-1/4x^4+C

Mar 1, 2018

int\ x^3ln(x^4)\ "d"x=4x^4ln(x)-x^4+C

Explanation:

Here's another approach:

Consider substituting first w=x^4 and ("d"w)/("d"x)=4x^3 into the integral int\ x^3ln(x^4)\ "d"x.

This gives 1/4int\ ln(w)\ "d"w.

Now use integration by parts int\ u\ "d"v=uv-int\ v\ "d"u. Use u=ln(w) and "d"v="d"w. Then find that "d"u=("d"w)/w and v=w.

Now, we have 1/4int\ ln(w)\ "d"w=1/4(wln(w)-int\ "d"w)
or simply 1/4wln(w)-1/4w+C.

Substitute back w=x^4.

Our final answer is thus 1/4x^4ln(x^4)-1/4x^4+C=x^4ln(x)-1/4x^4+C.